Take an equilateral triangle of side $1$.
Bisect it through a vertex and the midpoint of the opposite side.
You now have two right-angled triangles with angles $30^\circ, 60^\circ, 90^\circ$ with the edge opposite the $30^\circ$ of length $\frac12$ and hypotenuse $1$. So $$\sin (30^\circ)=\frac{\frac12}{1}=\frac12.$$
You can try this simpler alternative, which is in the spirit of finding the area of $\triangle GHB$ in terms of $x$.
Since we basically have a right triangle $ADE$, we can show that:
$$x^2+(2x)^2=DE^2\implies DE=\sqrt{x^2+5x^2}=x\sqrt5 \,\,\text{and:}\\
\sin \angle ADE=\frac{2x}{x\sqrt5}\implies \angle ADE=\arcsin\left(\frac2 {\sqrt{5}} \right)$$
It follows that $\angle CDG=90-\angle ADE$ and $\angle CDG=\angle DEA$.
Now we need to find $DG$:
$$\cos \angle CDG=\frac x{DG}\\
\cos\left(90-\arcsin\left(\frac2 {\sqrt{5}} \right)\right)=\sin\left(\arcsin\left(\frac2 {\sqrt{5}} \right)\right)=\frac x{DG}\\
\implies DG=\frac{x\sqrt5}{2}$$
It follows that:
$$x^2+CG^2=DG^2=\frac{5x^2}{4}\implies CG=\sqrt{\frac{x^2}{4}}=\frac{x}{2} \,\,\text{implying}:\\
GB=CG=\frac x2$$
Now we need to find $\angle DGC$, since $\angle DGC=\angle BGH$:
$$\sin \angle DGC= \frac{2x}{x\sqrt5}\implies \angle DGC=\angle BGH=\arcsin\left(\frac2 {\sqrt{5}} \right)$$
Since $\angle CBA$ is right, and $\angle FBE=60^\circ$, then it follows that $\angle GBH=30^\circ$
Now, since we know $\angle FBE=30^\circ$ and $\angle BEH=\angle DEA=90-\arcsin\left(\frac2 {\sqrt{5}} \right)$, then $\angle BHE$ is:
$$\angle BHE=30^\circ+\arcsin\left(\frac2 {\sqrt{5}} \right) \,\, \text{thus:}\\
\angle GHB=150^\circ -\arcsin\left(\frac2 {\sqrt{5}} \right)$$
Now, we need to find $BH$. By the law of sines, we have:
$$\frac{\sin\angle GHB}{GB}=\frac{\sin BGH}{HB}\,\,\,\, \text{implying:}\\
HB=\frac{GB\sin\angle BGH}{\sin\angle GHB}=\frac{\frac x2\sin\left(\arcsin\left(\frac2 {\sqrt{5}} \right)\right)}{\sin\left(150^\circ-\arcsin\left(\frac2 {\sqrt{5}} \right)\right)}$$
Keeping in mind that: $\cos(\arcsin(x))=\sqrt{1-x^2}$, we get that:
$$HB=\frac{2x}{1+2\sqrt3}$$
Now, finally, recalling that:
$$\triangle=\frac12 ab\sin A \,\,\text{where:}\\
$$
For $a:=GB=\frac x2,
b:=HB=\frac{2x}{1+2\sqrt3},
A:=\angle GBH=30^\circ$
Thus:
$$\therefore \triangle=\frac{x^2}{4(1+2\sqrt3)}$$
Best Answer
Alternatively, we can find that the triangle containing $D$ and $E$ has a base of $1$ and a height of $\frac{2}{3}$ (obtained by solving $1-x = \frac{1}{2} + \frac{1}{2}x$), so it has an area of $\frac{1}{2} \times 1 \times \frac{2}{3} = \frac{1}{3}$.
The height of $D$ is $\frac{2}{3} - \frac{1}{2} = \frac{1}{6}$, so the sides are $\frac{1}{4}$ of the big triangle. Since triangles $D$ and $D + E$ are similar by AA, the area of $D$ is $\frac{1}{16}$ times smaller. This gives the area of triangle $D$ as $\frac{1}{3} \times \frac{1}{16} = \frac{1}{48}$.